Improved Euler/Heun's Method - Differential Equations

1. One step of Heun's method is used to approximate $y(0.5)$ for the IVP $y' = 2y - 4x$, $y(0) = y_0$. With a step size of $h=0.5$, the approximation obtained is $y_1 = 3$. What was the value of the initial condition $y_0$?
   Answer: $1.4$ When you encounter Heun's method problems, remember that this is a two-stage Runge-Kutta method that uses both the slope at the current point and a predicted slope at the next point to make its approximation. For Heun's method with $y' = f(x,y)$, the formula is: $y_1 = y_0 + \frac{h}{2}[f(x_0, y_0) + f(x_0 + h, y_0 + hf(x_0, y_0))]$ Given $y' = 2y - 4x$ with $y(0) = y_0$, $h = 0.5$, and $y_1 = 3$: First, calculate $f(0, y_0) = 2y_0 - 4(0) = 2y_0$ Next, find the predictor: $y_0 + hf(0, y_0) = y_0 + 0.5(2y_0) = 2y_0$ Then calculate $f(0.5, 2y_0) = 2(2y_0) - 4(0.5) = 4y_0 - 2$ Applying Heun's formula: $3 = y_0 + \frac{0.5}{2}[2y_0 + (4y_0 - 2)]$ $3 = y_0 + 0.25[6y_0 - 2]$ $3 = y_0 + 1.5y_0 - 0.5$ $3 = 2.5y_0 - 0.5$ $3.5 = 2.5y_0$ $y_0 = 1.4$ Choice A (1.0) would give $y_1 = 2$, not 3. Choice B (1.5) results in $y_1 = 3.25$. Choice C (1.75) yields $y_1 = 3.625$. These all stem from computational errors in applying Heun's two-stage process. The answer is D. Study tip: Always work backwards systematically when given the final result in numerical methods problems. Set up the complete formula first, then substitute known values and solve algebraically.
2. For which of the following differential equations will one step of Euler's method and one step of Heun's method, starting from any initial condition $(x_0, y_0)$ and using the same step size $h$, produce the exact same approximation for $y(x_0+h)$?
   Answer: $y' = -10$ Let $y_1^E$ be the Euler approximation and $y_1^H$ be the Heun's approximation. $y_1^E = y_0 + h f(x_0, y_0)$. $y_1^H = y_0 + \frac{h}{2} [f(x_0, y_0) + f(x_0+h, y_1^*)]$, where $y_1^* = y_0 + h f(x_0, y_0) = y_1^E$. For the two approximations to be equal, we must have: $y_0 + h f(x_0, y_0) = y_0 + \frac{h}{2} [f(x_0, y_0) + f(x_0+h, y_1^E)]$ $h f(x_0, y_0) = \frac{h}{2} [f(x_0, y_0) + f(x_0+h, y_1^E)]$ $2 f(x_0, y_0) = f(x_0, y_0) + f(x_0+h, y_1^E)$ $f(x_0, y_0) = f(x_0+h, y_1^E)$. This condition means that the slope function $f(x,y)$ must have the same value at the initial point $(x_0, y_0)$ and at the predicted next point $(x_0+h, y_1^E)$. This is only guaranteed to be true for any $(x_0, y_0)$ and $h$ if the function $f(x,y)$ is a constant. For the choice $y'=-10$, we have $f(x,y) = -10$. Thus, $f(x_0, y_0) = -10$ and $f(x_0+h, y_1^E) = -10$. The condition is met.
3. For the IVP $\frac{dy}{dx} = \frac{x}{y}$, $y(2) = 3$, using Heun's method with $h = 0.4$, what is the most significant source of error in the first step compared to the analytical solution $y = \sqrt{x^2 + 5}$?
   Answer: The relatively large step size amplifies truncation errors from the trapezoidal approximation Heun's method has local truncation error $O(h^3)$, and the primary error source with $h = 0.4$ is that this step size is relatively large for the curvature of the solution. The trapezoidal rule approximation in the corrector step introduces significant error when $h$ is not small enough. Choice A misidentifies the linearization issue. Choice B incorrectly suggests over-compensation. Choice C is vague and doesn't identify the primary error mechanism.
4. When applying Heun's method to the IVP $y' = x + y^2$, $y(0)=1$ with a step size of $h=1$, a student correctly computes the predictor value $y_1^*=2$. What is the correct value for the corrector slope $m_2 = f(x_1, y_1^*)$?
   Answer: $5$ When you encounter Heun's method problems, remember that this predictor-corrector technique uses two slope calculations: an initial predictor slope and a corrector slope based on the predicted y-value. Given the differential equation $y' = x + y^2$ with $y(0) = 1$ and step size $h = 1$, you need to find the corrector slope $m_2 = f(x_1, y_1^*)$ where the student already found $y_1^* = 2$. First, identify your coordinates. Starting at $(x_0, y_0) = (0, 1)$, after one step of size $h = 1$, you reach $x_1 = 0 + 1 = 1$. The corrector slope is simply the function $f(x,y) = x + y^2$ evaluated at the point $(x_1, y_1^*) = (1, 2)$. Substituting these values: $m_2 = f(1, 2) = 1 + 2^2 = 1 + 4 = 5$. This confirms answer D is correct. Looking at the wrong answers: A) $1$ represents only the $x$ component, ignoring the $y^2$ term entirely. B) $4$ captures only the $y^2 = 2^2$ component while forgetting to add the $x = 1$ term. C) $2$ likely comes from mistakenly using the predicted $y$-value itself rather than evaluating the function. The key strategy for Heun's method problems is to carefully track your coordinates and systematically substitute into the given differential equation. Always double-check that you're evaluating $f(x,y)$ at the correct point—many errors stem from using wrong coordinates or forgetting terms in the function.
5. Consider the initial value problem $y' = x^2 - y$ with $y(0) = 1$. One step of the Improved Euler (Heun's) method with a step size of $h=0.1$ is used to find an approximation $y_1 \approx y(0.1)$. The method uses a predictor slope $m_1 = f(x_0, y_0)$ and a corrector slope $m_2 = f(x_1, y_1^*)$. What is the value of the corrector slope $m_2$?
   Answer: $-0.89$ The Improved Euler (Heun's) method consists of a predictor step and a corrector step. Given $f(x,y) = x^2 - y$, $x_0=0$, $y_0=1$, and $h=0.1$. 1. Calculate the predictor slope at the initial point: $m_1 = f(x_0, y_0) = f(0, 1) = 0^2 - 1 = -1$. 2. Use the predictor slope to find a preliminary estimate for $y_1$ (the predictor value $y_1^*$): $y_1^* = y_0 + h \cdot f(x_0, y_0) = 1 + 0.1 \cdot (-1) = 0.9$. 3. Calculate the corrector slope $m_2$ at the new point $(x_1, y_1^*) = (0.1, 0.9)$: $m_2 = f(x_1, y_1^*) = f(0.1, 0.9) = (0.1)^2 - 0.9 = 0.01 - 0.9 = -0.89$. The question asks for the value of the corrector slope $m_2$.
6. The Improved Euler (Heun's) method approximates the next value in a solution by using the average of two slopes. Geometrically, what do these two slopes represent?
   Answer: The slope at the beginning of the interval and an estimated slope at the end of the interval. Heun's method first calculates the slope at the current point $(x_n, y_n)$, which is $m_1 = f(x_n, y_n)$. It then uses this slope to make a preliminary estimate (a 'prediction') of the value at the end of the interval, $y_{n+1}^* = y_n + h \cdot m_1$. It then calculates a second slope, $m_2 = f(x_{n+1}, y_{n+1}^*)$, which is an estimate for the slope at the end of the interval. The final approximation is made using the average of these two slopes. Therefore, the two slopes are the slope at the beginning of the interval and an estimated slope at the end.
7. For the system $\frac{dy}{dx} = 2x + y$, $y(0) = 1$, suppose Heun's method with $h = 0.3$ gives $y(0.3) \approx 1.423$ while the analytical solution yields $y(0.3) = e^{0.3} - 2(0.3) - 2 \approx 1.350$. If the step size is halved to $h = 0.15$, approximately what value should Heun's method produce at $x = 0.3$?
   Answer: $y(0.3) \approx 1.368$ Heun's method has global error $O(h^2)$. The original error was $1.423 - 1.350 = 0.073$. When step size halves, error reduces by factor of $4$: new error $\approx 0.073/4 = 0.018$. So $y(0.3) \approx 1.350 + 0.018 = 1.368$. Choice B uses factor of $2$ reduction. Choice C uses $\sqrt{4}$ factor. Choice D assumes error reduces linearly with step size.
8. Consider applying Heun's method to the IVP $y' = -10y$, $y(0)=1$. The true solution, $y(x) = e^{-10x}$, decays toward zero. For the numerical approximation $y_1$ after one step to be stable and exhibit decay (i.e., $|y_1| < |y_0|$), the step size $h$ must be less than what maximum value?
   Answer: $0.2$ We need to find the condition on $h$ such that $|y_1| < |y_0|$. Here, $y_0=1$. First, let's compute $y_1$ in terms of $h$. $f(x,y) = -10y$, $(x_0, y_0) = (0,1)$. $m_1 = f(0,1) = -10(1) = -10$. $y_1^* = y_0 + h m_1 = 1 - 10h$. $m_2 = f(h, y_1^*) = -10(1-10h) = -10 + 100h$. $y_1 = y_0 + \frac{h}{2}(m_1 + m_2) = 1 + \frac{h}{2}(-10 - 10 + 100h) = 1 + \frac{h}{2}(-20 + 100h) = 1 - 10h + 50h^2$. The condition for stability is $|y_1| < 1$, which translates to $-1 < 1 - 10h + 50h^2 < 1$. First inequality: $1 - 10h + 50h^2 < 1 \implies 50h^2 - 10h < 0 \implies 10h(5h-1) < 0$. Since $h>0$, we need $5h-1 < 0$, which gives $h < 1/5$ or $h < 0.2$. Second inequality: $-1 < 1 - 10h + 50h^2 \implies 0 < 2 - 10h + 50h^2$. The quadratic $50h^2 - 10h + 2$ has discriminant $\Delta = (-10)^2 - 4(50)(2) = 100 - 400 = -300 < 0$. Since the leading coefficient is positive, the quadratic is always positive. Thus, this inequality holds for all $h$. Combining the conditions, the step size must satisfy $h < 0.2$.
9. Use Heun's method with a step size of $h=0.2$ to approximate the solution of the initial value problem $y' = 4 - y^2$, $y(0)=0$ at $x=0.4$.
   Answer: $1.278$ This problem requires two steps of Heun's method. Let $f(x,y)=4-y^2$. Step 1: Approximate $y(0.2)$ from $(x_0, y_0) = (0,0)$ with $h=0.2$. $m_1 = f(0,0) = 4 - 0^2 = 4$. $y_1^* = y_0 + h m_1 = 0 + 0.2(4) = 0.8$. $m_2 = f(x_1, y_1^*) = f(0.2, 0.8) = 4 - (0.8)^2 = 4 - 0.64 = 3.36$. $y_1 = y_0 + \frac{h}{2}(m_1+m_2) = 0 + \frac{0.2}{2}(4+3.36) = 0.1(7.36) = 0.736$. So $y(0.2) \approx 0.736$. Step 2: Approximate $y(0.4)$ from $(x_1, y_1) = (0.2, 0.736)$ with $h=0.2$. $m_1' = f(0.2, 0.736) = 4 - (0.736)^2 = 4 - 0.541696 = 3.458304$. $y_2^* = y_1 + h m_1' = 0.736 + 0.2(3.458304) = 0.736 + 0.6916608 = 1.4276608$. $m_2' = f(x_2, y_2^*) = f(0.4, 1.4276608) = 4 - (1.4276608)^2 \approx 4 - 2.03821 = 1.96179$. $y_2 = y_1 + \frac{h}{2}(m_1'+m_2') \approx 0.736 + \frac{0.2}{2}(3.4583 + 1.9618) = 0.736 + 0.1(5.4201) = 0.736 + 0.54201 = 1.27801$. The closest answer is $1.278$.
10. The local truncation error of Heun's method is on the order of $O(h^3)$. This is achieved by constructing the approximation $y_{n+1}$ such that its Taylor series expansion in powers of $h$ matches the Taylor series expansion of the true solution $y(x_n+h)$. Up to which power of $h$ do these two expansions match?
    Answer: $h^2$ The order of a numerical method is related to its local truncation error. The local truncation error is defined as the difference between the true solution and the numerical approximation after one step, assuming the previous value was exact. For a method to have a local truncation error of order $O(h^{p+1})$, its Taylor series expansion must match the true solution's Taylor series expansion up to and including the term of order $h^p$. Since Heun's method has a local truncation error of $O(h^3)$, its expansion must match the true solution's expansion up to the $h^2$ term. The mismatch first occurs at the $h^3$ term, which defines the order of the local error. Therefore, the expansions match for terms with powers $h^0, h^1$, and $h^2$.
11. A student is approximating the solution to $y' = f(x,y)$ with $y(x_n)=y_n$ and step size $h$. They perform the following two calculations: $k_1 = h \cdot f(x_n, y_n)$ $y_{n+1} = y_n + \frac{1}{2} (k_1 + h \cdot f(x_n+h, y_n+k_1))$ This calculation scheme represents which numerical method?
    Answer: Improved Euler (Heun's) Method Let's analyze the given formulas. The first calculation is $k_1 = h \cdot f(x_n, y_n)$. This represents the increment in $y$ according to Euler's method. The term $y_n+k_1$ is therefore $y_n + h \cdot f(x_n, y_n)$, which is the predictor step in Heun's method, often denoted $y_{n+1}^*$. Substituting this into the second equation: $y_{n+1} = y_n + \frac{1}{2} (h f(x_n, y_n) + h f(x_n+h, y_{n+1}^*))$ Factoring out $h/2$: $y_{n+1} = y_n + \frac{h}{2} [f(x_n, y_n) + f(x_n+h, y_{n+1}^*)]$. This is the standard formula for the Improved Euler (Heun's) method, which averages the slope at the beginning of the interval, $f(x_n, y_n)$, and the estimated slope at the end of the interval, $f(x_n+h, y_{n+1}^*)$.
12. When applying Heun's method to $\frac{dy}{dx} = -2y + 3x$, $y(1) = 2$, with step size $h = 0.25$, which expression correctly represents the predictor step from $x = 1$ to $x = 1.25$?
    Answer: $y_1^* = 2 + 0.25[-2(2) + 3(1)]$ In Heun's method, the predictor step uses only the slope at the initial point: $y_1^* = y_0 + h \cdot f(x_0, y_0)$. Here, $f(1,2) = -2(2) + 3(1) = -1$, so $y_1^* = 2 + 0.25(-1)$. Choice B incorrectly uses the endpoint $x$-value in the predictor. Choice C represents the corrector step formula. Choice D attempts to average slopes in the predictor step, which is incorrect procedure.
13. For $\frac{dy}{dx} = \frac{1}{1+x^2} - \frac{y}{2}$, $y(0) = 0$, Heun's method with $h = 0.4$ produces $y(0.4) \approx 0.2847$. The analytical solution is $y(x) = 2[\arctan(x) - \frac{x}{1+x^2} + Ce^{-x/2}]$ where $C$ satisfies the initial condition. What is the primary reason for any discrepancy between numerical and analytical results?
    Answer: Heun's method accumulates truncation error from the trapezoidal approximation over the integration interval The primary source of error in Heun's method is the local truncation error $O(h^3)$ at each step, which accumulates to global error $O(h^2)$. This comes from the trapezoidal rule approximation inherent in the method. Choice A is irrelevant to numerical vs analytical comparison. Choice C misidentifies the error source. Choice D overestimates the impact of round-off error compared to truncation error for this step size.
14. Apply one step of Heun's method to the initial value problem $y' = \frac{2x}{y+1}$, $y(1)=1$ with a step size of $h=0.5$. What is the resulting approximation for $y(1.5)$?
    Answer: $1.550$ We apply one step of Heun's method with $f(x,y) = \frac{2x}{y+1}$, $(x_0, y_0) = (1,1)$, and $h=0.5$. 1. Calculate the predictor slope $m_1$: $m_1 = f(1,1) = \frac{2(1)}{1+1} = 1$. 2. Calculate the predictor value $y_1^*$: $y_1^* = y_0 + h \cdot m_1 = 1 + 0.5(1) = 1.5$. 3. Calculate the corrector slope $m_2$ at $(x_1, y_1^*) = (1.5, 1.5)$: $m_2 = f(1.5, 1.5) = \frac{2(1.5)}{1.5+1} = \frac{3}{2.5} = \frac{6}{5} = 1.2$. 4. Calculate the final approximation $y_1$: $y_1 = y_0 + \frac{h}{2}(m_1+m_2) = 1 + \frac{0.5}{2}(1+1.2) = 1 + 0.25(2.2) = 1 + 0.55 = 1.55$.
15. Using Heun's method with step size $h = 0.2$, approximate $y(0.4)$ for the initial value problem $\frac{dy}{dx} = x + y$, $y(0) = 1$. What is the value obtained after applying the predictor-corrector sequence twice?
    Answer: $y(0.4) \approx 1.5936$ Heun's method uses predictor-corrector: First step: $y_1^* = 1 + 0.2(0 + 1) = 1.2$, then $y_1 = 1 + 0.1[(0 + 1) + (0.2 + 1.2)] = 1.24$. Second step: $y_2^* = 1.24 + 0.2(0.2 + 1.24) = 1.528$, then $y_2 = 1.24 + 0.1[(0.2 + 1.24) + (0.4 + 1.528)] = 1.5936$. Choice A uses only the predictor value. Choice C incorrectly adds an extra correction term. Choice D uses forward Euler throughout.
16. Two students solve $\frac{dy}{dx} = y - x^2$, $y(1) = 2$, using $h = 0.25$. Student A uses Heun's method properly. Student B mistakenly applies the corrector formula twice in each step (i.e., uses the first corrector result as input to compute a second corrector). At $x = 1.25$, which relationship holds?
    Answer: Student B's result has larger error than Student A's due to over-correction instability Applying the corrector formula multiple times in Heun's method leads to over-correction and can cause instability, especially when the step size is not very small. The method is designed for one predictor and one corrector step. Multiple corrections don't improve accuracy and typically introduce additional error. Choice A is incorrect as over-correction worsens accuracy. Choice B is wrong since extra correction does change the result. Choice D misunderstands that this doesn't create a higher-order method.
17. Consider applying Heun's method to $\frac{dy}{dx} = \sin(xy)$, $y(0) = \pi/6$, with $h = 0.1$. After computing the predictor $y_1^*$, a student finds that $f(0.1, y_1^*)$ requires evaluating $\sin(0.1 \cdot y_1^*)$. If $y_1^* \approx 0.524$, what is the most appropriate way to handle the corrector calculation?
    Answer: Use calculator or numerical evaluation of $\sin(0.0524)$ as the method requires Heun's method requires accurate evaluation of the function $f(x,y)$ at both predictor and corrector stages. Using approximations (choices A, D) or unnecessary series expansions (choice B) defeats the purpose of the numerical method. Standard numerical evaluation is appropriate and necessary. Choice A introduces additional error. Choice B is unnecessarily complex. Choice D changes the differential equation being solved.
18. Consider the initial value problem $y' = 2x - 3$, with $y(1) = 5$. If one step of Heun's method with step size $h$ is used to approximate $y(1+h)$, what is the local error of this approximation?
    Answer: $0$ Heun's method has a local truncation error of $O(h^3)$, meaning it can find the exact value for polynomial solutions of degree 2 or less in a single step. First, find the exact solution to the IVP by integration: $y(x) = \int (2x - 3) dx = x^2 - 3x + C$. Using $y(1)=5$, we have $5 = 1^2 - 3(1) + C \implies 5 = -2 + C \implies C=7$. The exact solution is $y(x) = x^2 - 3x + 7$. The exact value at $x=1+h$ is $y(1+h) = (1+h)^2 - 3(1+h) + 7 = (1+2h+h^2) - (3+3h) + 7 = h^2 - h + 5$. Now, apply one step of Heun's method from $(x_0, y_0) = (1,5)$: $f(x,y) = 2x-3$. $m_1 = f(1, 5) = 2(1) - 3 = -1$. Predictor: $y_1^* = y_0 + h \cdot m_1 = 5 + h(-1) = 5-h$. Corrector slope: $m_2 = f(1+h, 5-h) = 2(1+h) - 3 = 2+2h-3 = 2h-1$. Approximation: $y_1 = y_0 + \frac{h}{2}(m_1 + m_2) = 5 + \frac{h}{2}(-1 + 2h-1) = 5 + \frac{h}{2}(2h-2) = 5 + h^2 - h$. The error is the exact value minus the approximation: $(h^2 - h + 5) - (5 + h^2 - h) = 0$. The approximation is exact.

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